Problem: Simplify and expand the following expression: $ \dfrac{2y}{5y - 5}-\dfrac{2y - 3}{3y - 8} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(5y - 5)(3y - 8)$ Multiply the first term by $\dfrac{3y - 8}{3y - 8}$ $ \begin{align*} \dfrac{2y}{5y - 5} \times \dfrac{3y - 8}{3y - 8} & = \dfrac{(2y)(3y - 8)}{(5y - 5)(3y - 8)} \\ & = \dfrac{6y^2 - 16y}{(5y - 5)(3y - 8)}\end{align*} $ Multiply the second term by $\dfrac{5y - 5}{5y - 5}$ $ \begin{align*} \dfrac{2y - 3}{3y - 8} \times \dfrac{5y - 5}{5y - 5} & = \dfrac{(2y - 3)(5y - 5)}{(3y - 8)(5y - 5)} \\ & = \dfrac{10y^2 - 25y + 15}{(3y - 8)(5y - 5)}\end{align*} $ Now we have: $ = \dfrac{6y^2 - 16y}{(5y - 5)(3y - 8)} - \dfrac{10y^2 - 25y + 15}{(3y - 8)(5y - 5)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{6y^2 - 16y - (10y^2 - 25y + 15)}{(5y - 5)(3y - 8)} $ $ = \dfrac{6y^2 - 16y - 10y^2 + 25y - 15}{(5y - 5)(3y - 8)} $ $ = \dfrac{-4y^2 + 9y - 15}{(5y - 5)(3y - 8)}$ Expand the denominator: $ = \dfrac{-4y^2 + 9y - 15}{15y^2 - 55y + 40}$